Quadratic number fields containing primitive roots of unity

Question

A problem from Michael Artin's Algebra (Second Edition) from Fields: Determine the quadratic number fields $\mathbb{Q}[\sqrt{d}]$ that contain a primitive $n$th root of unity, for some integer $n$.

I do not know how to get about. Any hints?

Answer 1

Let $\alpha$ be a primitive $n$th root of unity for some $n$, such that $\alpha$ is contained in a quadratic extension. So the minimal polynomial of $\alpha$ is of degree $2$, meaning that $(x-\alpha)(x-\overline{\alpha})\in\mathbb{Q}[x]$. It is a known fact that the $n$th cyclotomic polynomial is irreducible, thus it follows that $\phi(n)=2$, where $\phi$ is Euler's phi function. It is not hard to find all possible values of $n$ that satisfy this.

Answer 2

The degree of a primitive $n$-th root of unity as an algebraic number over $\mathbb{Q}$ is $\varphi(n)$, hence $\exp\left(\frac{2\pi i}{n}\right)$ may belong to a quadratic number field only if $\varphi(n)=2$, i.e. for $n=3$ ($\mathbb{Q}[\sqrt{-3}]$) and $n=4$ ($\mathbb{Q}[\sqrt{-1}]$).

Answer 3

For variety,

For every $n$ that doesn't divide $24$, there exists a residue class $x$ modulo $n$ that is relatively prime to $n$ and such that $x^2 \neq 1 \pmod{n}$. Therefore, there exists a prime number $p$ such that $n$ does not divide $p^2 - 1$.

If $K$ is a field containing a primitive $n$-th root of unity, then for every prime $p$ not dividing $n$, all of its residue fields of characteristic $p$ have a primitive $n$-th root of unity $\zeta$. If we choose $p$ such that $p^2 \not \equiv 1 \pmod n$, then neither $\mathrm{GF}(p)$ nor its quadratic extension has a primitive $n$-th root of unity, and therefore $\mathrm{GF}(p)(\zeta)$ is an extension of $\mathrm{GF}(p)$ degree greater than $2$.

Therefore, $K$ must be an extension of $\mathbb{Q}$ of degree greater than $2$.

Thus, we need only consider which $\mathbb{Q}(\zeta)$ are quadratic fields, for $\zeta$ a primitive $1,2,3,4,6,8,12,24$-th root of unity.